Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.

Guest Apr 26, 2022

#1**+1 **

That question has been solved here: https://web2.0calc.com/questions/algebra-question_67

Griff Apr 26, 2022

#2**+1 **

(a^2)(b^3)=(20)^18.

20^18 =2^36 * 5^18

a =(2^18)^2 =262,144^2

b =(5^6)^3 =15,625^3

ryderbellamy Apr 26, 2022

#3**+1 **

20^18==2.62144e+23 = 2^36 * 5^18

Total number of prime factors= 54

**Total number of ordered pairs= 2,701**

20^18==2.62144e+23 = 2^36 * 5^18

Total number of prime factors= 54

**Total number of ordered triplets= 4,104,919**

Guest Apr 26, 2022

#4**+1 **

a=262,144

b=15,625

LCM[15,625, 262,144)=11,790,000

11790000 = 2^4 * 3^2 * 5^4 * 131

**Number of ordered pairs=[ (4*2+1) + (2*2+1) + (4*2+1) + (2*1+1)]==9 * 5 * 9 *3==1,215**

Guest Apr 26, 2022

#5**+1 **

Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.

\( 20^{18}=2^{36}*5^{18}\)

\(5^0*5^{18}\\ 5^6*5^{12}\\ \\~\\ 2^0*2^{36}\\ 2^6*2^{30}\\ 2^{12}*2^{24}\\ 2^{18}*2^{18}\\ \)

5^0 can be paired with 7 of them (I mean different powers of 2)

5^6 can be paired with 7 of them

So I just get 14 but order counts, and a and b are all interchangeable so that is 28

Here they are

If you can spot any that I have missed then please let me know.

(I am not guarenteeing that this answer is right)

Melody Apr 27, 2022